2= (>>60) (z<80) 5. A The ages of the thousands of residents of a retirement community are normally distributed with a mean of 70 and a standard deviation of 4 years. a. What proportion of this population is between 60 and 80? X.4 2= 2=60-70 = -2.5 = 6.21 4 2-80-76 J = 2.5=6.21 1-,99379 -3 = 6.21% +6.2 12.24% is bett 60- Year: b. If one sample of 45 residents is chosen at random, what is the probability that the sample mean age will be between 68.5 and 71.75? 68.5-45 2= 25.87
2= (>>60) (z<80) 5. A The ages of the thousands of residents of a retirement community are normally distributed with a mean of 70 and a standard deviation of 4 years. a. What proportion of this population is between 60 and 80? X.4 2= 2=60-70 = -2.5 = 6.21 4 2-80-76 J = 2.5=6.21 1-,99379 -3 = 6.21% +6.2 12.24% is bett 60- Year: b. If one sample of 45 residents is chosen at random, what is the probability that the sample mean age will be between 68.5 and 71.75? 68.5-45 2= 25.87
MATLAB: An Introduction with Applications
6th Edition
ISBN:9781119256830
Author:Amos Gilat
Publisher:Amos Gilat
Chapter1: Starting With Matlab
Section: Chapter Questions
Problem 1P
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![**Transcription for Educational Website**
---
**Problem 5**
A: The ages of the thousands of residents of a retirement community are normally distributed with a mean of 70 and a standard deviation of 4 years.
**a. What proportion of this population is between 60 and 80?**
\[
Z = \frac{X - \mu}{\sigma}
\]
- For age 60:
\[
Z = \frac{60 - 70}{4} = -2.5 \approx -6.21
\]
- For age 80:
\[
Z = \frac{80 - 70}{4} = 2.5 \approx 6.21
\]
Calculation:
1 - (Probability of Z < -6.21 + Probability of Z > 6.21) = 0.99379
Proportion = 6.21% + 6.21% = 12.24%
**Conclusion:** 12.24% of the population is between 60 and 80 years old.
---
**b. If one sample of 45 residents is chosen at random, what is the probability that the sample mean age will be between 68.5 and 71.75?**
\[
Z = \frac{\bar{X} - \mu}{\sigma/\sqrt{n}}
\]
- For age 68.5:
\[
Z = \frac{68.5 - 70}{4/\sqrt{45}} = -5.87
\]
- For age 71.75:
\[
Z = \frac{71.75 - 70}{4/\sqrt{45}} = 6.6
\]
Diagrams and detailed calculations show that this configuration results in a probability of a sample mean being in this range, with students marked -3 for incorrect graph plotting and -4 for calculation errors.
**Note:** Exact Z scores, such as -6.21, indicate more precise computations and rounding in context.](/v2/_next/image?url=https%3A%2F%2Fcontent.bartleby.com%2Fqna-images%2Fquestion%2F1f106ccb-8482-4a66-9989-84251f794043%2F990a5d9d-6f71-4f08-8311-82f02806996b%2Fmgduube_processed.jpeg&w=3840&q=75)
Transcribed Image Text:**Transcription for Educational Website**
---
**Problem 5**
A: The ages of the thousands of residents of a retirement community are normally distributed with a mean of 70 and a standard deviation of 4 years.
**a. What proportion of this population is between 60 and 80?**
\[
Z = \frac{X - \mu}{\sigma}
\]
- For age 60:
\[
Z = \frac{60 - 70}{4} = -2.5 \approx -6.21
\]
- For age 80:
\[
Z = \frac{80 - 70}{4} = 2.5 \approx 6.21
\]
Calculation:
1 - (Probability of Z < -6.21 + Probability of Z > 6.21) = 0.99379
Proportion = 6.21% + 6.21% = 12.24%
**Conclusion:** 12.24% of the population is between 60 and 80 years old.
---
**b. If one sample of 45 residents is chosen at random, what is the probability that the sample mean age will be between 68.5 and 71.75?**
\[
Z = \frac{\bar{X} - \mu}{\sigma/\sqrt{n}}
\]
- For age 68.5:
\[
Z = \frac{68.5 - 70}{4/\sqrt{45}} = -5.87
\]
- For age 71.75:
\[
Z = \frac{71.75 - 70}{4/\sqrt{45}} = 6.6
\]
Diagrams and detailed calculations show that this configuration results in a probability of a sample mean being in this range, with students marked -3 for incorrect graph plotting and -4 for calculation errors.
**Note:** Exact Z scores, such as -6.21, indicate more precise computations and rounding in context.
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